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AD-5: Aufgabe 6 bearbeitet.
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Jim Martens
@ -99,4 +99,145 @@ Jim Martens (6420323)}
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\subsection{} %a
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\subsection{} %b
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\section{} %6
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\begin{tikzpicture}[shorten >=1pt,node distance=1.1cm]
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\node[state] (s) {s};
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\node[state] (a) [above right=1.5 and 1.0 of s] {A};
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\node[state] (b) [below right=1.5 and 1.0 of s] {B};
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\node[state] (c) [above right=0.5 and 1.0 of b] {C};
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\node[state] (d) [below right=0.5 and 1.0 of b] {D};
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\node[state] (e) [above right=0.5 and 1.0 of a] {E};
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\node[state] (f) [below right=0.5 and 1.0 of a] {F};
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\node[state] (g) [above right=0.5 and 2.0 of c] {G};
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\path[every node/.style={font=\scriptsize},->]
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(s) edge node [above] {30} (a)
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(s) edge node [below] {18} (b)
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(a) edge node [above] {20} (e)
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(a) edge node [below] {22} (f)
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(b) edge node [above] {21} (c)
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(b) edge node [below] {38} (d)
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(f) edge node [above] {9} (g)
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(c) edge node [below] {9} (g);
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\end{tikzpicture}
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Da dies ein gerichteter Graph ohne Zyklen ist und auch keine negativen Kantengewichte vorkommen, kann der Dijkstra-Algorithmus angewendet werden. Da von 9 bis 17 Uhr Fahrer bezahlt werden müssen, kommen nur die Pfade in Betrachtung, die zu einer Senke führen. Senken sind D, E und G. Derjenige dieser Knoten welcher als erster vom Algorithmus gefunden wird (über den eindeutig bestimmbaren Weg), ist auch der kürzeste Pfad vom Startknoten aus, der die Bedingungen erfüllt.
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Es ergibt sich folgende Abfolge:
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\begin{verbatim}
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S = {s}
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d(s) = 0
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while 1.
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U = {A, B}
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for all u in U -> u = A
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for all pre(u) in S that are predecessors of u -> pre(u) = s
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d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
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d'(u, pre(u)) = 0 + 30
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for all u in U -> u = B
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for all pre(u) in S that are predecessors of u -> pre(u) = s
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d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
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d'(u, pre(u)) = 0 + 18
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u* = B
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d(u*) = 18
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S = {s, B}
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while 2.
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U = {A, C, D}
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for all u in U -> u = A
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for all pre(u) in S that are predecessors of u -> pre(u) = s
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d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
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d'(u, pre(u)) = 0 + 30
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for all u in U -> u = C
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for all pre(u) in S that are predecessors of u -> pre(u) = B
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d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
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d'(u, pre(u)) = 18 + 21
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for all u in U -> u = D
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for all pre(u) in S that are predecessors of u -> pre(u) = B
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d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
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d'(u, pre(u)) = 18 + 38
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u* = A
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d(u*) = 30
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S = {S, B, A}
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while 3.
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U = {C, D, E, F}
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for all u in U -> u = C
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for all pre(u) in S that are predecessors of u -> pre(u) = B
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d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
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d'(u, pre(u)) = 18 + 21
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for all u in U -> u = D
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for all pre(u) in S that are predecessors of u -> pre(u) = B
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d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
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d'(u, pre(u)) = 18 + 38
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for all u in U -> u = E
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for all pre(u) in S that are predecessors of u -> pre(u) = A
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d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
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d'(u, pre(u)) = 30 + 20
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for all u in U -> u = F
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for all pre(u) in S that are predecessors of u -> pre(u) = A
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d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
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d'(u, pre(u)) = 30 + 22
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u* = C
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d(u*) = 39
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S = {s, B, A, C}
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while 4.
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U = {D, E, F, G}
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for all u in U -> u = D
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for all pre(u) in S that are predecessors of u -> pre(u) = B
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d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
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d'(u, pre(u)) = 18 + 38
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for all u in U -> u = E
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for all pre(u) in S that are predecessors of u -> pre(u) = A
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d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
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d'(u, pre(u)) = 30 + 20
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for all u in U -> u = F
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for all pre(u) in S that are predecessors of u -> pre(u) = A
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d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
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d'(u, pre(u)) = 30 + 22
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for all u in U -> u = G
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for all pre(u) in S that are predecessors of u -> pre(u) = C
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d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
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d'(u, pre(u)) = 39 + 9
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u* = G
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d(u*) = 48
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S = {s, B, A, C, G}
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while 5.
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U = {D, E, F}
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for all u in U -> u = D
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for all pre(u) in S that are predecessors of u -> pre(u) = B
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d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
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d'(u, pre(u)) = 18 + 38
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for all u in U -> u = E
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for all pre(u) in S that are predecessors of u -> pre(u) = A
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d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
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d'(u, pre(u)) = 30 + 20
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for all u in U -> u = F
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for all pre(u) in S that are predecessors of u -> pre(u) = A
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d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
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d'(u, pre(u)) = 30 + 22
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u* = E
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d(u*) = 50
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S = {s, B, A, C, G, E}
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while 6.
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U = {D, F}
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for all u in U -> u = D
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for all pre(u) in S that are predecessors of u -> pre(u) = B
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d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
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d'(u, pre(u)) = 18 + 38
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for all u in U -> u = F
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for all pre(u) in S that are predecessors of u -> pre(u) = A
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d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
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d'(u, pre(u)) = 30 + 22
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u* = F
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d(u*) = 52
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S = {s, B, A, C, G, E, F}
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while 7.
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U = {D}
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for all u in U -> u = D
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for all pre(u) in S that are predecessors of u -> pre(u) = B
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d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
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d'(u, pre(u)) = 18 + 38
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u* = D
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d(u*) = 56
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S = {s, B, A, C, G, E, F, D}
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\end{verbatim}
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Wie zu sehen ist wird die Senke G zuerst erreicht. Folgt man dem Weg zu G, so ergibt sich, dass der kürzeste Pfad von s über B und C nach G führt. Weniger Kosten als 48 sind daher nicht möglich.
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\end{document}
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